HW 4 SOLUTIONS
Q1
What is the mole fraction of vanillin (Mm = 152.1 g/mol) in an aqueous solution that is 3.65 m?
SOLN:
First solve for moles of vanillin, using molality. Then determine mole fraction, converting the
kilogram of water to moles of water.
mv=
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
Xv = 𝑛𝑛
Q2
𝑘𝑘𝑘𝑘 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑛𝑛 𝑣𝑣
𝑣𝑣 +𝑛𝑛 𝑤𝑤
=
=
𝑛𝑛 𝑣𝑣
𝑘𝑘𝑘𝑘
→ 𝑛𝑛v = 3.65 moles
3.65 𝑚𝑚𝑚𝑚𝑚𝑚
3.65 𝑚𝑚𝑚𝑚𝑚𝑚 +(1𝑘𝑘𝑘𝑘 𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤𝑤 )�
1000 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚
��
�
𝑘𝑘𝑘𝑘
18 𝑔𝑔
= 0.0616
What is the molarity of NaCl in an aqueous solution that is 8.90 % mass in NaCl? The density of
the solution is 1.075 g/mL.
SOLN:
We are given mass%, but with conversion factors, we can determine the concentration. Change
the numerator to moles and the denominator to liters.
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀% =
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚𝑚𝑚(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
→ 𝑀𝑀 =
𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝐿𝐿(𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠)
8.90
𝑚𝑚𝑚𝑚𝑚𝑚
100 𝑔𝑔 �58.44𝑔𝑔�
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀 =
= 1.64 𝑀𝑀
𝑚𝑚𝑚𝑚
𝐿𝐿
1𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 �
��
�
1.075𝑔𝑔 1000𝑚𝑚𝑚𝑚
Q3
To what volume should 33 mL of a 11.2 M HCl solution be diluted in order to prepare a 0.49 M
HCl solution?
SOLN:
Use the molarity equation twice.
𝑚𝑚𝑚𝑚𝑚𝑚
→ 𝑛𝑛 = 𝑀𝑀𝑀𝑀 = (11.2𝑀𝑀)(0.033𝐿𝐿) = 0.3696𝑚𝑚𝑚𝑚𝑚𝑚
𝐿𝐿
𝑛𝑛 0.3696 𝑚𝑚𝑚𝑚𝑚𝑚
𝑉𝑉 = =
= 0.754 𝑚𝑚𝑚𝑚 = 750 𝑚𝑚𝑚𝑚
𝑀𝑀
0.49 𝑀𝑀
𝑀𝑀 =
Q4
The following successive dilutions are applied to a stock solution that is 1.80 M sucrose
•
•
•
Solution A = 48.0 mL of the stock solution is diluted to 178. mL
Solution B = 97.9 mL of Solution A is diluted to 218. mL
Solution C = 89.0 mL of Solution B is diluted to 269. mL
What is the concentration of sucrose in solution C?SOLN:
There are 3 dilutions, therefore, there should be 3 dilution factors. Start with the initial molarity
and dilute 3 times.
𝑉𝑉
𝑉𝑉
𝑉𝑉
48
97.9
89
𝑀𝑀𝐶𝐶 = 𝑀𝑀0 �𝑉𝑉0 � �𝑉𝑉𝐴𝐴 � �𝑉𝑉𝐵𝐵 � = 1.80 𝑀𝑀 �178 � � 218 � �269 �= 0.0721 M
𝐴𝐴
𝐵𝐵
𝐶𝐶
Q5
How many L of 3.70 M HCl are required to react completely with 13.0 g of Al(OH)3 ?
Al(OH)3 + 3HCl AlCl3 + 3H2O
SOLN:
Convert from g to moles of Al(OH)3, then use stoichiometry to see how many moles of HCl are
required.
Then use the molarity equation.
13 𝑔𝑔 𝐴𝐴𝐴𝐴(𝑂𝑂𝑂𝑂)3 �
𝑉𝑉 =
Q6
𝑚𝑚𝑚𝑚𝑚𝑚
3 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻𝐻𝐻𝐻𝐻
��
� = 0.5 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻𝐻𝐻𝐻𝐻
36.46 𝑔𝑔 𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴(𝑂𝑂𝑂𝑂)3
𝑛𝑛 0.5𝑚𝑚𝑚𝑚𝑚𝑚
=
= 0.135 𝐿𝐿
𝑀𝑀
3.7𝑀𝑀
A 43.30 mL of sample of Ca(OH)2 solution is titrated with 0.2758 M HCl. What is the molarity of
the calcium hydroxide solution if the titration requires 69.90 mL of the acid to reach the
endpoint?
SOLN:
𝑛𝑛 = (0.06990 𝐿𝐿)(0.2758 𝑀𝑀 𝐻𝐻𝐻𝐻𝐻𝐻) �
𝑀𝑀 =
1 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶(𝑂𝑂𝑂𝑂)2
� = 0.964 𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶(𝑂𝑂𝑂𝑂)2
2 𝑚𝑚𝑚𝑚𝑚𝑚 𝐻𝐻𝐻𝐻𝐻𝐻
𝑛𝑛
0.964 𝑚𝑚𝑚𝑚𝑚𝑚
=
= 0.2226 𝑀𝑀
𝑉𝑉 0.0433 + 0.0699 𝐿𝐿
Q7
Complete a reaction table in millimoles for the reaction of 63.9 mL of 0.9900 M Ag1+ and 43.8
mL of 1.0500 M CrO42-.
2Ag1+ + CrO42Ag2CrO4
63.3
46.0
0
initial
mmol
-63.3
-31.6
31.6
delta
mmol
0.0
final
14.4
31.6
mmol
What mass in grams of Ag2CrO4 forms?
10.48
What is the molar concentration of the remaining excess reactant? Assume that the volumes are
additive.
0.133
SOLN:
First fill in the table: For all initial values, multiply the volume (in mL) by the concentration (in
M) to get millimoles. 𝑛𝑛 = (63.9𝑚𝑚𝑚𝑚)(0.99𝑀𝑀) = 63.3 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 Ag+
Silver ions are the limiting reagent since: 63.3 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 �
1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶2𝑂𝑂4
2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
� = 31.6 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶𝐶𝐶4
Therefore, 31.6 mmol are used up of CrO42- and all of Ag+. To determine the product formed,
use stoichiometry.
1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴2𝐶𝐶𝐶𝐶𝐶𝐶4
331.74 𝑚𝑚𝑚𝑚
��
� = 10.5 𝑔𝑔
2 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴
1 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴2𝐶𝐶𝐶𝐶𝐶𝐶4
14.4 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐶𝐶𝐶𝐶𝐶𝐶4
𝑀𝑀 =
= 0.1337𝑀𝑀
43.8 + 63.9 𝑚𝑚𝑚𝑚
63.3 𝑚𝑚𝑚𝑚𝑚𝑚𝑚𝑚 𝐴𝐴𝐴𝐴 �
Q8
What is the molarity of KBr in an aqueous solution that is 5% mass in KBr? The density of the
solution is 1.06 g/mL.
SOLN:
To get from mass% to molarity, use conversion factors of molecular mass and density.
5 𝑔𝑔 𝐾𝐾𝐾𝐾𝑟𝑟
𝑀𝑀𝑀𝑀𝑀𝑀𝑀𝑀% =
100 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
1.06 𝑔𝑔 1000 𝑚𝑚𝑚𝑚
𝑚𝑚𝑚𝑚𝑚𝑚
5 𝑔𝑔 𝐾𝐾𝐾𝐾𝐾𝐾
��
��
��
� = 0.445 𝑀𝑀
𝑀𝑀 = �
100 𝑔𝑔 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠
𝑚𝑚𝑚𝑚
𝐿𝐿
119 𝑔𝑔